Problem: The $n\text{th}$ partial sum of the series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ is given by ${{S}_{n}}=6-\frac{3}{{{n}^{2}}}$. $\sum\limits_{n=1}^{9}{{{a}_{n}}}=$
Explanation: $\sum\limits_{n=1}^{9}{{{a}_{n}}}~$ is the same as ${{S}_{9}}\,$. ${{S}_{9}}=6-\frac{3}{{{9}^{2}}}=6-\frac{1}{27}=\frac{161}{27}$